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Ta có \(\overrightarrow{IB}=\overrightarrow{BA}\Rightarrow\hept{\begin{cases}I\in AB\\\overrightarrow{AI}=2\overrightarrow{AB}\end{cases}}\). Tương tự \(\hept{\begin{cases}J\in\left[AC\right]\\\overrightarrow{AJ}=\frac{AJ}{AC}\overrightarrow{AC}=\frac{2}{5}\overrightarrow{AC}\end{cases}}\)
Do đó \(\overrightarrow{IJ}=\overrightarrow{AJ}-\overrightarrow{AI}=\frac{2}{5}\overrightarrow{AC}-2\overrightarrow{AB}\)(đpcm).
giải giúp t câu này nha : tính vecto IG theo vecto AB và vecto AC (các b vẽ hình ra hộ t nhé)
a: \(\overrightarrow{AE}=\dfrac{2}{3}\overrightarrow{EC}\)
=>E nằm giữa A và C và AE=2/3EC
Ta có: AE+EC=AC(E nằm giữa A và C)
=>\(AC=\dfrac{2}{3}EC+EC=\dfrac{5}{3}EC\)
=>\(\dfrac{AE}{AC}=\dfrac{\dfrac{2}{3}EC}{\dfrac{5}{3}EC}=\dfrac{2}{3}:\dfrac{5}{3}=\dfrac{2}{5}\)
=>\(AE=\dfrac{2}{5}AC\)
=>\(\overrightarrow{AE}=\dfrac{2}{5}\cdot\overrightarrow{AC}\)
\(\overrightarrow{BE}=\overrightarrow{BA}+\overrightarrow{AE}\)
\(=-\overrightarrow{AB}+\dfrac{2}{5}\cdot\overrightarrow{AC}\)
b: \(\left|\overrightarrow{IA}+\overrightarrow{IG}\right|=\left|\overrightarrow{IA}-\overrightarrow{IG}\right|\)
=>\(\left[{}\begin{matrix}\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IA}-\overrightarrow{IG}\\\overrightarrow{IA}+\overrightarrow{IG}=\overrightarrow{IG}-\overrightarrow{IA}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}2\cdot\overrightarrow{IG}=\overrightarrow{0}\\2\cdot\overrightarrow{IA}=\overrightarrow{0}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}I\equiv G\\I\equiv A\end{matrix}\right.\)
\(5\overrightarrow{JB}=2\overrightarrow{JC}=2\left(\overrightarrow{JB}+\overrightarrow{BC}\right)=2\overrightarrow{JB}+2\overrightarrow{BC}\)
\(\Rightarrow\overrightarrow{JB}=\dfrac{2}{3}\overrightarrow{BC}=2\overrightarrow{BA}+2\overrightarrow{AC}\Rightarrow\overrightarrow{BJ}=2\overrightarrow{AB}-2\overrightarrow{AC}\)
\(\Rightarrow\overrightarrow{AJ}=\overrightarrow{AB}+\overrightarrow{BJ}=\overrightarrow{AB}+2\overrightarrow{AB}-2\overrightarrow{AC}=3\overrightarrow{AB}-2\overrightarrow{AC}\)
A B C G I J
a) \(\overrightarrow{IA}+2\overrightarrow{IB}=\overrightarrow{BA}+3\overrightarrow{IB}=\overrightarrow{0}\Rightarrow\overrightarrow{BI}=\frac{1}{3}\overrightarrow{BA}\)
\(\overrightarrow{CI}=\overrightarrow{CB}+\overrightarrow{BI}=\overrightarrow{CB}+\frac{1}{3}\overrightarrow{BA}=\overrightarrow{CB}+\frac{1}{3}\left(\overrightarrow{CA}-\overrightarrow{CB}\right)=\frac{2}{3}\overrightarrow{CB}+\frac{1}{3}\overrightarrow{CA}\)
\(\overrightarrow{JB}=x\overrightarrow{JC}\Rightarrow\overrightarrow{CB}-\overrightarrow{CJ}=x\overrightarrow{JC}\Rightarrow\overrightarrow{CB}=\left(x-1\right)\overrightarrow{JC}\Rightarrow\overrightarrow{CJ}=\frac{1}{1-x}\overrightarrow{CB}\)
b) \(\overrightarrow{IJ}=\overrightarrow{CJ}-\overrightarrow{CI}=\frac{1}{1-x}\overrightarrow{CB}-\left(\frac{2}{3}\overrightarrow{CB}+\frac{1}{3}\overrightarrow{CA}\right)=\frac{2x+1}{3\left(1-x\right)}\overrightarrow{CB}-\frac{1}{3}\overrightarrow{CA}\)
c) Dễ có \(\overrightarrow{CG}=\frac{2}{3}\left(\overrightarrow{CB}+\overrightarrow{CA}\right)\). Để \(\overrightarrow{IJ}\)//\(\overrightarrow{CG}\) thì :
\(\frac{\frac{2}{3}}{\frac{2x+1}{3\left(1-x\right)}}=\frac{\frac{2}{3}}{-\frac{1}{3}}\Leftrightarrow\frac{1-x}{2x+1}=-1\Rightarrow2x+1=x-1\Leftrightarrow x=-2\)
Vậy \(x=-2\)tức \(\overrightarrow{JB}=-2\overrightarrow{JC}\)thì IJ // CG.
* Nhận xét: Nếu \(\overrightarrow{u}=x\overrightarrow{a}+y\overrightarrow{b};\overrightarrow{v}=m\overrightarrow{a}+n\overrightarrow{b}\)thì \(\overrightarrow{u}\)//\(\overrightarrow{v}\)\(\Leftrightarrow\frac{x}{m}=\frac{y}{n}.\)